sorry i didn't respond quicker but i found out i hadn't uploaded my db to the server :( sorry for wasting your time "Richard Hutchins" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > Haven't seen that this has been solved yet so I thought I'd throw my two > cents into the fray. First of all, I'd second the earlier suggestion that > you assign your query to a variable. And, second, I'd recommend that you > echo out your query to the browser just before it is submitted to the > database server. That way, you can actually _see_ what's being sent to the > server. I have found this to be an invaluable troubleshooting tactic and it > may help you find the root cause of your invalid MySQL result. > > > > -----Original Message----- > > From: David Robley [mailto:[EMAIL PROTECTED] > > Sent: Wednesday, June 23, 2004 2:53 AM > > To: [EMAIL PROTECTED] > > Subject: Re: [PHP-DB] problem.... > > > > > > Water_foul wrote: > > > > > i fixed those things and it didn't fix it :( :( :( :( :( :( > > :( :( :( is > > > there a icq chatroom for php?) > > > "Shahmat Bin Dahlan" <[EMAIL PROTECTED]> wrote in message > > > news:[EMAIL PROTECTED] > > >> Your SQL statement: > > >> > > >> 'SELECT Rune, username FROM RuneRunner > > >> RuneRunner_1 WHERE (User_ID = 3)' > > >> > > >> What is "RuneRunner" and "RuneRunner_1"? Are these two > > different tables. > > >> If it is, you might want to separate them with a comma. > > >> > > >> Why not try getting rid of the brackets surrounding > > "User_ID=3"? And > > >> also wrap single quotes around the digit "3". > > >> > > >> > > >> ----- Original Message ----- > > >> From: "water_foul" <[EMAIL PROTECTED]> > > >> Date: Wednesday, June 23, 2004 8:58 am > > >> Subject: Re: [PHP-DB] problem.... > > >> > > >> > I checked em all they were right > > >> > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message > > >> > > > news:[EMAIL PROTECTED] > > >> > > Sounds like it doesn't like your SQL statement. > > Perhaps a field > > >> > or table > > >> > > name is incorrect? > > >> > > > > >> > > > Warning: mysql_fetch_array(): supplied argument is > > not a valid > > >> > MySQL> > result > > >> > > > resource in ....... > > >> > > > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message > > >> > > > > > >> > > > news:[EMAIL PROTECTED]> > > >> >> What error are you getting? > > >> > > >> > > >> > > >> > why doesn't this work: > > >> > > >> > $pic=mysql_query('SELECT Rune, username FROM > > >> > RuneRunner> >> > RuneRunner_1 WHERE (User_ID = > > 3)',$connection); > > >> > > >> > $pic=mysql_fetch_array($pic); > > >> > > > > > Two suggestions for debugging mysql problems: > > > > 1) Create your query as a variable so you can echo it and see > > exactly what > > is being passed to mysql > > > > $query = "SELECT Rune, username FROM RuneRunner, RuneRunner_1 > > WHERE User_ID > > = 3"; > > > > 2) Use mysql_error() to get the actual error from mysql > > $result = mysql_query($query,$connection); > > echo mysql_error(); > > $pic=mysql_fetch_array($result); > > > > -- > > David Robley > > > > "I'd like to learn a new card game," Tom said wistfully. > > > > -- > > PHP Database Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > >
-- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
